Let $s\in(0,1),$ $\Omega\subset\mathbb{R}^{N}$ regular and limited, $v\in H_{0}^{s}(\Omega)$ and $u:\Omega\rightarrow\mathbb{R},$ such that, \begin{equation} \iint\limits_{\Omega\times\Omega}\frac{|u(x)-u(y)|^{2}}{|x-y|^{N+2s}}dxdy<+\infty. \end{equation} My question is:
If $0\leq u\leq v$ can we say that $u\in H_{0}^{s}(\Omega)$?
Here is the full proof. First, since $v\in L^2$, so is $u$ and so $u\in H^1(\Omega)$.
So the question is about the behavior near the boundary. But warning about the fact that if $s\leq 1/2$, then $H^s_0=H^s$ (see e.g. L. Tartar, 2007, "An Introduction to Sobolev spaces and Interpolation Spaces"). So in this case this is true.
Another simple case is if $s>N/2$, then by Sobolev embeddings (more precisely Morrey inequality), $u$ and $v$ are continuous and therefore since $v$ converges to $0$ at the boundary pointwise, so is $u$ by your inequality.
The missing cases are when $N≥2$ and $s\in(1/2,N/2]$. Then one can characterize $H^1_0(\Omega)$ as the kernel of the operator of trace at the boundary $\gamma_0$ which is a bounded operator $H^{s}(\Omega)\to H^{s-1/2}(\partial\Omega)$, defined as the extension of the mapping of evaluation at the boundary defined on the dense subset of continuous functions (see again e.g. Tartar, 2007). Now extend the functions $u$ and $v$ to $H^s(\mathbb R^N)$ using an usual continuous extension operator. Then mollifying the functions (i.e. taking $u_n=\varphi_n*u$ with $\varphi_n=n^d\varphi(nx)∈C^\infty_c$ nonnegative and with integral $1$), then $u_n$ and $v_n$ converge to $u$ and $v$ in $H^s$ and one keeps the property that $0≤ u_n≤v_n$. Since these functions are continuous, the trace is just the evaluation on the boundary and so $0≤ \gamma_0(u_n)≤\gamma_0(v_n)$ a.e. on $\partial\Omega$. By continuity of the trace and of the extension operator, one can pass to the limit to get $0≤ \gamma_0(u)≤\gamma_0(v)$ a.e. on $\partial\Omega$. Since $\gamma_0(v)=0$ on $\partial\Omega$, we deduce that the same holds for $u$ and so $u∈H^s_0(\Omega)$.